Basic Probability

Probability is a measure that is associated with how certain we are of results, or outcomes, of a particular activity. When the activity is a planned operation carried out under controlled conditions, it is called an experiment. If the result is not predetermined, then the experiment is said to be a chance experiment. Each time the experiment is attempted is called a trial.

Examples of chance experiments include the following:

• flipping a fair coin,
• spinning a spinner,
• drawing a marble at random from a bag, and
• rolling a pair of dice.
  

A result of an experiment is called an outcome. The sample space of an experiment is the set, or collection, of all possible outcomes.

There are four main ways to represent a sample space:

$0 ≤ P(A) ≤ 1.P(A) = 0$

means the event A can never happen. $P(A) = 1$ means the event $A$ always happens.

$P(A) = 0.5$

means the event A is equally likely to occur or not to occur.

Figure 3.6

If two outcomes or events are equally likely, then they have equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. This is known as the theoretical probability of A.

Theoretical Probability of Event A

$P(A)=\dfrac{\text{Number of outcomes in event A}}{\text{Total number of possible outcomes}}$.

For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. Let A represent the outcome getting one head. There are two outcomes that meet this condition {HT, TH}, so

$P(A)=\dfrac{2}{4}=\dfrac{1}{2}=.5$.

Theoretical probability is not sufficient in all situations, however. Suppose we want to calculate the probability that a randomly selected car will run a red light at a given intersection. In this case, we need to look at events that have occurred, not theoretical possibilities. We could install a traffic camera and count the number of times that cars failed to stop when the light was red and the total number of cars that passed through the intersection for a period of time. These data will allow us to calculate the experimental, or empirical, probability that a car runs the red light.

Experimental Probability of Event A

$P(A)=\dfrac{\text{Number of times event A occurs}}{\text{Total number of trials}}$

While theoretical and experimental methods provide two different ways to calculate probability, these methods are closely related. If you flip one fair coin, there is one way to obtain heads and two possible outcomes. So, the theoretical probability of heads is $\dfrac{1}{2}$. Probability does not predict short-term results, however. If an experiment involves flipping a coin 10 times, you should not expect exactly five heads and five tails. The probability of any outcome measures the long-term relative frequency of that outcome. If you continue to flip the coin (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches .5 (the probability of heads). This important characteristic of probability experiments is known as the law of large numbers, which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed, or empirical, relative frequency will approach the theoretical probability.

Suppose you roll one fair, six-sided die with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}.

$P(E)=\dfrac{2}{6}$.

If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, $\dfrac{2}{6}$ of the rolls would result in an outcome of at least five. You would not expect exactly $\dfrac{2}{6}$, but the long-term relative frequency of obtaining this result would approach the theoretical probability of $\dfrac{2}{6}$ as the number of repetitions grows larger and larger.

It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one-euro coin and discovered that in 250 trials, a head was obtained 56 percent of the time and a tail was obtained 44 percent of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.

OR Event

An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let $A = {1, 2, 3, 4, 5}$ and $B = {4, 5, 6, 7, 8}$. $A \text{ OR }B = {1, 2, 3, 4, 5, 6, 7, 8}$. Notice that 4 and 5 are not listed twice.

AND Event

An outcome is in the event $A \text { AND }B$ if the outcome is in both A and B at the same time. For example, let $A \text {and} B$ be
${1, 2, 3, 4, 5}$ and ${4, 5, 6, 7, 8}$, respectively. Then $A \text{ AND }B = {4, 5}$.

The complement of event A is denoted A' (read "A prime"). A' consists of all outcomes that are not in A. Notice that

$P(A) + P(A') = 1$. For example, let $S = {1, 2, 3, 4, 5, 6}$ and let $A = {1, 2, 3, 4}$. Then, $A' = {5, 6}$. $P(A) = \dfrac{4}{6}$, $P(A') = \dfrac{2}{6}$, and $P(A) + P(A') = \dfrac{4}{6}+\dfrac{2}{6}= 1$.

The conditional probability of $A$ given $B$ is written $P(A|B)$, read "the probability of $A$, given $B$". $P(A|B)$ is the probability that event $A$ will occur given that the event $B$ has already occurred. A conditional probability reduces the sample space. We calculate the probability of $A$ from the reduced sample space $B$. The formula to calculate $P(A|B)$ is $P(A|B) =\dfrac{P(A AND B)}{P(B)}$ where $P(B)$ is greater than zero.

For example, suppose we toss one fair, six-sided die. The sample space $S = {1, 2, 3, 4, 5, 6}$. Let $A = {2, 3}$ and $B = {2, 4, 6}$. $P(A|B)$ represents the probability that a randomly selected outcome is in $A$ given that it is in $B$. We know that the outcome must lie in $B$, so there are three possible outcomes. There is only one outcome in $B$ that also lies in $A$, so $P(A|B) = \dfrac{1}{3}$.

We get the same result by using the formula. Remember that S has six outcomes.

$P(A|B) = \dfrac{P(A \text{ AND }B)}{P(B)}=\dfrac{\dfrac{\text{(the number of outcomes that are 2 or 3 and even in S)}}{6}}{\dfrac{\text{(the number of outcomes that are even in S)}}{6}}=\dfrac{\dfrac{1}{6}}{\dfrac{3}{6}}=\dfrac{1}{3}$

Understanding Terminology and Symbols

It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.

Example 3.1

The sample space S is the whole numbers starting at one and less than 20.

1. $S =$ ________
2. Let event $A =$ the even numbers and event $B =$ numbers greater than 13.
3. $A =$ ________, $B =$ ________
4. $P(A) =$ ________, $P(B) =$ ________
5. $A \text{ AND }B =$ ________, $A \text{ OR }B =$ ________
6. $P(A \text{ AND }B) =$ ________, $P(A \text{ OR }B) =$ ________
7. $A' =$ ________, $P(A') =$ ________
8. $P(A) + P(A') =$ ________
9. $P(A|B) =$ ________, $P(B|A) =$ ________; are the probabilities equal?

Solution

1. $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}$
   
   
2. $A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19}$
   
   
3. $P(A) = \dfrac{\text{number of outcomes in A}} {\text{number of outcomes in S}} = \dfrac{9}{19}$, $P(B) = \dfrac{\text{number of outcomes in B}}{\text{number of outcomes in S}} = \dfrac{6}{19}$
   
   
4. The set $A \text{ AND }B$ contains all outcomes that lie in both sets $A$ and $B$, so $A \text{ AND } B = {14,16,18}$, The set $A \text{ OR }B$ contains all outcomes that lie either of the sets $A$ or $B$, so $A \text{ OR }B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}$.
   
   
5. $P(A \text{ AND }B) = \dfrac{3}{19}$, $P(A \text{ OR }B) = \dfrac{12}{19}$
   
   
6. A' consists of all outcomes in the sample space, $S$, that DO NOT lie in $A$, so $A' = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P(A') =\dfrac{10}{19}$.
   
   
7. $P(A) + P(A') = \dfrac{9}{19} + \dfrac{10}{19} = 1$
   
   
8. $P(A|B) = \dfrac{P(A\text { AND }B)}{P(B)} =\dfrac{\dfrac{3}{19}}{\dfrac{6}{19}} = \dfrac{3}{6}$, $P(B|A) = \dfrac{P(A\text{ AND }B)}{P(A)} =\dfrac{\dfrac{3}{19}}{\dfrac{9}{19}} = \dfrac{3}{9}$, No, the probabilities are not equal.
   
   

Try It 3.1

The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

1. $S =$ ________
   Let event $A =$ the sum is even and event $B =$ the first number is prime.
2. $A =$ ________, $B =$ ________
3. $P(A) =$ ________, $P(B) =$ ________
4. $A \text{ AND }B =$ ________, $A \text{ OR }B =$ ________
5. $P(A \text{ AND }B) =$ ________, $P(A \text{ OR }B) =$ ________
6. $B' =$ ________, $P(B') =$ ________
7. $P(A) + P(A') =$ ________
8. $P(A|B) =$ ________, $P(B|A) =$ ________; are the probabilities equal?

Example 3.2

A fair, six-sided die is rolled. The sample space, $S$, is {1, 2, 3, 4, 5, 6}. Describe each event and calculate its probability.

1. Event $T =$ the outcome is two.
2. Event $A =$ the outcome is an even number.
3. Event $B =$ the outcome is less than four.
4. The complement of $A$
5. $A$ GIVEN $B$
6. $B$ GIVEN $A$
7. $A$ AND $B$
8. $A$ OR $B$
9. $A$ OR $B'$
10. Event $N =$ the outcome is a prime number.
11. Event $I =$ the outcome is seven.

Solution

1. $T = {2}$, $P(T) = \dfrac{\text{number of outcomes in T}}{\text {number of outcomes in S}} =\dfrac{1}{6}$
2. $A = {2, 4, 6}$, $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$
3. $B = {1, 2, 3}$, $P(B) = \dfrac{3}{6} = \dfrac{1}{2}$
4. $A' = {1, 3, 5}$, $P(A') = \dfrac{3}{6} = \dfrac{1}{2}$
5. $A|B = {2}$, There are three outcomes in $B$, and only 1 of these lies in $A$, so $P(A|B) = \dfrac{1}{3}$
   
6. $B|A = {2}$, There are three outcomes in $A$, and only 1 of these lies in $B$, so $P(B|A) = \dfrac{1}{3}$
7. $A\text{ AND }B = {2}$, $P(A \text{ AND }B) = \dfrac{1}{6}$
8. $A \text{ OR }B = {1, 2, 3, 4, 6}$, $P(A \text{ OR }B) = \dfrac{5}{6}$
9. $A \text{ OR }B' = {2, 4, 5, 6}$, $P(A \text{ OR }B') =\dfrac{4}{6} = \dfrac{2}{3}$
10. $N = {2, 3, 5}$, $P(N) = \dfrac{1}{2}$
11. It is impossible to roll a die and get an outcome of 7, so $P(7) = 07$.

Example 3.3

Table 3.2 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right or left-handed.

1. $P(M)$
2. $P(F)$
3. $P(R)$
4. $P(L)$
5. $P(M \text{ AND }R)$
6. $P(F \text{ AND }L)$
7. $P(M \text{ OR }F)$
8. $P(M \text{ OR }R)$
9. $P(F \text{ OR }L)$
10. $P(M')$
11. $P(R|M)$
12. $P(F|L)$
13. $P(L|F)$

Solution

1. $P(M)=\dfrac{\text{ number of males}} {\text{ total number of subjects}}=\dfrac{43+9}{43+9+44+4}=\dfrac{52}{100}=.52$
   
   
2. $P(F)=\dfrac{\text{ number of females}}{\text{total number of subjects}}=\dfrac{44+4}{43+9+44+4}=\dfrac{48}{100}=.48$
   
   
3. $P(R)=\dfrac{\text{number of right-handed subjects}}{\text{total number of subjects}}=\dfrac{43+44}{43+9+44+4}=\dfrac{87}{100}=.87$
   
   
4. $P(L)=\dfrac{\text{number of left-handed subjects}}{\text{total number of subjects}}=\dfrac{9+4}{43+9+44+4}=\dfrac{13}{100}=.13$
   
   
5. $P(M \text{ and }R)=\dfrac{\text{number of male, right-handed subjects}}{\text{total number of subjects}}=\dfrac{43}{100}=.43$
   
   
6. $P(F \text { and } L)=\dfrac{\text{number of female, left-handed subjects}}{\text{total number of subjects}}=\dfrac{4}{100}=.04$
   
   
7. $P(M\text{ or }F)=\dfrac{\text{number of subjects that are male or female}}{\text{total number of subjects}}=\dfrac{52+48}{100}=\dfrac{100}{100}=1$
   
   
8. $P(M\text{ or} R)=\dfrac{\text{ number of subjects that are male or right-handed}}{\text{ total number of subjects}}=\dfrac{43+9+44}{100}=\dfrac{96}{100}=.96$
   
   
9. $P(F \text{ or }L)=\dfrac{\text{number of subjects that are female or left-handed}}{\text{total number of subjects}}=\dfrac{44+4+9}{100}=\dfrac{57}{100}=.57$
   
   
10. $P(M')=\dfrac{\text{number of subjects who are not male}}{\text{total number of subjects}}=\dfrac{44+4}{43+9+44+4}=\dfrac{48}{100}=.48$
    
    
11. $P(R | M)= \dfrac{P(R\text { and }M)}{P(M)}=\dfrac{0.43}{0.52}˜=.8269$
    (rounded to four decimal places)
    
    
12. $P(F | L)=\dfrac{P(F\text{ and }L)}{P(L)}=\dfrac{0.04}{0.13}˜=.3077$
    (rounded to four decimal places)
    
    
13. $DP(L | F)=\dfrac{P(L\text{ and }F)}{P(F)}=\dfrac{0.04}{0.48}˜=.0833$
    (rounded to four decimal places)

Source: OpenStax, https://openstax.org/books/statistics/pages/3-introduction
This work is licensed under a Creative Commons Attribution 4.0 License.

Independent and mutually exclusive do not mean the same thing.

Independent Events

Two events are independent if the following are true:

• P( A| B) = P( A)
• P( B| A) = P( B)
• P( A AND B) = P( A) P( B)

Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are not independent, then we say that they are dependent events.

Sampling may be done with replacement or without replacement.

• With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.

A bag contains four blue and three white marbles. James draws one marble from the bag at random, records the color, and replaces the marble. The probability of drawing blue is $\dfrac{4}{7}$. When James draws a marble from the bag a second time, the probability of drawing blue is still $\dfrac{4}{7}$. James replaced the marble after the first draw, so there are still four blue and three white marbles.

• Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

The bag still contains four blue and three white marbles. Maria draws one marble from the bag at random, records the color, and sets the marble aside. The probability of drawing blue on the first draw is $\dfrac{4}{7}$. Suppose Maria draws a blue marble and sets it aside. When she draws a marble from the bag a second time, there are now three blue and three white marbles. So, the probability of drawing blue is now $\dfrac{3}{6} = \dfrac{1}{2}$. Removing the first marble without replacing it influences the probabilities on the second draw.

If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.

Example 3.4

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. Clubs and spades are black, while diamonds and hearts are red cards. There are 13 cards in each suit consisting of A (ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.

Figure 3.9

a. Sampling with replacement Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the 10 of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are { Q of spades, 10 of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck.

b. Sampling without replacement Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are { K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.

Try It 3.4

The Multiplication Rule

The Addition Rule

Example 3.14

• Klaus can only afford one vacation. The probability that he chooses A is $P(A) = .6$ and the probability that he chooses B is $P(B) = .35$.
• $P(A \text{ AND }B) = 0$ because Klaus can only afford to take one vacation.
• Therefore, the probability that he chooses either New Zealand or Alaska is $P(A \text{ OR }B) = P(A) + P(B) = .6 + .35 = .95$. Note that the probability that he does not choose to go anywhere on vacation must be .05.

Example 3.15

Solution 1

Solution 2

Solution 3

Solution 4

Try It 3.15

Example 3.16

Solution 1

Solution 3

Solution 4

Solution 5

Try It 3.16

Example 3.17

1. What is the probability that Felicity enrolls in math and speech?
2. Find P(M AND S) = P(M|S)P(S).
3. What is the probability that Felicity enrolls in math or speech classes?
4. Find P(M OR S) = P(M) + P(S) − P(M AND S).
5. Are M and S independent? Is P(M|S) = P(M)?
6. Are M and S mutually exclusive? Is P(M AND S) = 0?

Solution 1

Try It 3.17

1. Find P(B AND D).
2. Find P(B OR D).

Example 3.18

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Try It 3.18

Example 3.19

1. Given that a woman develops the disease, what is the probability that she tests positive? Find P(P|B) = 1 − P(N|B).
2. What is the probability that a woman develops the disease and tests positive? Find P(B AND P) = P(P|B)P(B).
3. What is the probability that a woman does not develop the disease? Find P(B′) = 1 − P(B).
4. What is the probability that a woman tests positive for the disease? Find P(P) = 1 − P(N).

Solution 1

Try It 3.19

1. Find P(B′).
2. Find P(D AND B).
3. Find P(B|D).
4. Find P(D AND B′).
5. Find P(D|B′).

In calculating probability, there are two rules to consider when you are determining if two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication Rule

If A and B are two events defined on a sample space, then P( A AND B) = P( B) P( A| B).

This equation can be rewritten as P( A AND B) = P( B) P( A| B), the multiplication rule.

If A and B are independent, then P( A| B) = P( A). In this special case, P( A AND B) = P( A| B) P( B) becomes P( A AND B) = P( A) P( B).

The Addition Rule

If A and B are defined on a sample space, then P( A OR B) = P( A) + P( B) − P( A AND B).

Draw one card from a standard deck of playing cards. Let H = the card is a heart, and let J = the card is a jack. These events are not mutually exclusive because a card can be both a heart and a jack.

$\text{P(H or J)=P(H)+P(J)−P(H and J)}$

$=\dfrac{13}{52}+\dfrac{4}{52}−\dfrac{1}{52}$

$=\dfrac{16}{52}$

$=\dfrac{4}{13}$

$≈.3077$

If A and B are mutually exclusive, then P( A AND B) = 0. Then P( A OR B) = P( A) + P( B) − P( A AND B) becomes P( A OR B) = P( A) + P( B).

Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is

$\text{P(H or S) = P(H)+P(S)}$

$=\dfrac{13}{52}+\dfrac{13}{52}$

$=\dfrac{26}{52}$

$=\dfrac{1}{2}$

$=.5$

Example 3.14

Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska.

• Klaus can only afford one vacation. The probability that he chooses A is P( A) = .6 and the probability that he chooses B is P( B) = .35.
• P( A AND B) = 0 because Klaus can only afford to take one vacation.
• Therefore, the probability that he chooses either New Zealand or Alaska is P( A OR B) = P( A) + P( B) = .6 + .35 = .95. Note that the probability that he does not choose to go anywhere on vacation must be .05.
  
  

Example 3.15

Carlos plays college soccer. He makes a goal 65 percent of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P( A) = .65. B = the event Carlos is successful on his second attempt. P( B) = .65. Carlos tends to shoot in streaks. The probability that he makes the second goal given that he made the first goal is .90.

a. What is the probability that he makes both goals?

Solution 1

a. The problem is asking you to find P( A AND B) = P( B AND A). Since P( B| A) = .90: P( B AND A) = P( B| A) P( A) = (.90)(.65) = .585.

Carlos makes the first and second goals with probability .585.

b. What is the probability that Carlos makes either the first goal or the second goal?

Solution 2

b. The problem is asking you to find P( A OR B).

P( A OR B) = P( A) + P( B) − P( A AND B) = .65 + .65 − .585 = .715

Carlos makes either the first goal or the second goal with probability .715.

c. Are A and B independent?

Solution 3

c. No, they are not, because P( B AND A) = .585.

P( B) P( A) = (.65)(.65) = .423

.423 ≠ .585 = P( B AND A)

So, P( B AND A) is not equal to P( B) P( A).

d. Are A and B mutually exclusive?

Solution 4

d. No, they are not because P( A and B) = .585.

To be mutually exclusive, P( A AND B) must equal zero.

Try It 3.15

Example 3.16

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

a. What is the probability that the member is a novice swimmer?

Solution 1

a. There are 150 members; 75 of these are advanced, and 47 of these are intermediate swimmers. So there are 150 − 75 − 47 = 28 novice swimmers. The probability that a randomly selected swimmer is a novice is $\dfrac{28}{150}$.

b. What is the probability that the member practices four times a week?

Solution 2

b. $\dfrac{40 + 30 + 10}{150}=\dfrac{80}{150}$

c. What is the probability that the member is an advanced swimmer and practices four times a week?

Solution 3

c. There are 40 advanced swimmers who practice four times per week, so the probability is $\dfrac{40}{150}$.

d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and being an intermediate swimmer mutually exclusive? Why or why not?

Solution 4

d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?

Solution 5

e. No, these are not independent events. P(novice AND practices four times per week) = .0667 P(novice) P(practices four times per week) = .0996 .0667 ≠ .0996

Try It 3.16

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their schools sports teams. What is the probability that a senior is taking a gap year?

Example 3.17

Felicity attends a school in Modesto, CA. The probability that Felicity enrolls in a math class is .2 and the probability that she enrolls in a speech class is .65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is .25.

Let M = math class, S = speech class, and M| S = math given speech.

1. What is the probability that Felicity enrolls in math and speech? Find P( M AND S) = P( M| S) P( S).
2. What is the probability that Felicity enrolls in math or speech classes? Find P( M OR S) = P( M) + P( S) − P( M AND S).
3. Are M and S independent? Is P( M| S) = P( M)?
4. Are M and S mutually exclusive? Is P( M AND S) = 0?

Solution 1

a. P( M AND S) = P( M| S) P( S) = .25(.65) = .1625

b. P( M OR S) = P( M) + P( S) − P( M AND S) = .2 + .65 − .1625 = .6875

c. No, P( M| S) = .25 and P( M) = .2.

d. No, P( M AND S) = .1625.

Try It 3.17

A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P( B) = .40, P( D) = .30, and P( D| B) = .5.

1. Find P( B AND D).
2. Find P( B OR D).
   
   

Example 3.18

Researchers are studying one particular type of disease that affects women more often than men. Studies show that about one woman in seven (approximately 14.3 percent) who live to be 90 will develop the disease. Suppose that of those women who develop this disease, a test is negative 2 percent of the time. Also suppose that in the general population of women, the test for the disease is negative about 85 percent of the time. Let B = woman develops the disease and let N = tests negative. Suppose one woman is selected at random.

a. What is the probability that the woman develops the disease? What is the probability that woman tests negative?

Solution 1

a. P( B) = .143; P( N) = .85

b. Given that the woman develops the disease, what is the probability that she tests negative?

Solution 2

b. Among women who develop the disease, the test is negative 2 percent of the time, so P( N| B) = .02

c. What is the probability that the woman has the disease AND tests negative?

Solution 3

c. P( B AND N) = P( B) P( N| B) = (.143)(.02) = .0029

d. What is the probability that the woman has the disease OR tests negative?

Solution 4

d. P( B OR N) = P( B) + P( N) − P( B AND N) = .143 + .85 − .0029 = .9901

e. Are having the disease and testing negative independent events?

Solution 5

e. No. P( N) = .85; P( N| B) = .02. So, P( N| B) does not equal P( N).

f. Are having the disease and testing negative mutually exclusive?

Solution 6

f. No. P( B AND N) = .0029. For B and N to be mutually exclusive, P( B AND N) must be zero.

Try It 3.18

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their school's sports teams. What is the probability that a senior is going to college and plays sports?

Example 3.19

Refer to the information in Example 3.18. P = tests positive.

1. Given that a woman develops the disease, what is the probability that she tests positive? Find P( P| B) = 1 − P( N| B).
2. What is the probability that a woman develops the disease and tests positive? Find P( B AND P) = P( P| B) P( B).
3. What is the probability that a woman does not develop the disease? Find P( B′) = 1 − P( B).
4. What is the probability that a woman tests positive for the disease? Find P( P) = 1 − P( N).

Solution 1

a. P( P| B) = 1 − P( N| B) = 1 − .02 = .98

b. P( B AND P) = P( P| B) P( B) = .98(.143) = .1401

c. P( B') = 1 − P( B) = 1 − .143 = .857

d. P( P) = 1 − P( N) = 1 − .85 = .15

Try It 3.19

A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P( B) = .40, P( D) = .30, and P( D| B) = .5.

1. Find P( B′).
2. Find P( D AND B).
3. Find P( B| D).
4. Find P( D AND B′).
5. Find P( D| B′).